Solve the following Question.(1 Marks)

Question 11 Mark

A function f is defined as follows:
f(x) = 5 – x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3.

Answer

f(x) = 5 – x
f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

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Question 21 Mark

If $f(x)=2 x^2+3, g(x)=5 x-2$, then find : gog

Answer

$\begin{aligned} & (g \circ g)(x)=g(g(x)) \\ & =g(5 x-2) \\ & =5(5 x-2)-2 \\ & =25 x-10-2 \\ & =25 x-12\end{aligned}$

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Question 31 Mark

If $f(x)=2 x^2+3, g(x)=5 x-2$, then find : fof

Answer

$\begin{aligned} &(f \circ f)(x)=f(f(x)) \\ & =f\left(2 x^2+3\right) \\ & =2\left(2 x^2+3\right)^2+3 \\ & =2\left(4 x^4+12 x^2+9\right)+3 \\ & =8 x^4+24 x^2+18+3 \\ & =8 x^4+24 x^2+21\end{aligned}$

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Question 41 Mark

If $f(x)=2 x^2+3, g(x)=5 x-2$, then find : gof

Answer

$\begin{aligned} & (g \circ f)(x)=g(f(x)) \\ = & g\left(2 x^2+3\right) \\ = & 5\left(2 x^2+3\right)-2 \\ = & 10 x^2+15-2 \\ = & 10 x^2+13\end{aligned}$

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Question 51 Mark

If $f(x)=2 x^2+3, g(x)=5 x-2$, then find : fog

Answer

$\begin{aligned} & (f \circ g)(x)=f(g(x)) \\ = & f(5 x-2) \\ = & 2(5 x-2)^2+3 \\ = & 2\left(25 x^2-20 x+4\right)+3 \\ = & 50 x^2-40 x+8+3 \\ = & 50 x^2-40 x+11\end{aligned}$

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Question 61 Mark

Find $x$, if $g(x)=0$ where : $g(x)=6 x^2+x-2$

Answer

$\begin{aligned} & g(x)=6 x^2+x-2 \\ & g(x)=0 \\ & \therefore 6 x^2+x-2=0 \\ & \therefore 6 x^2+4 x-3 x-2=0 \\ & \therefore 2 x(3 x+2)-1(3 x+2)=0 \\ & \therefore(2 x-1)(3 x+2)=0 \\ & \therefore 2 x-1=0 \text { or } 3 x+2=0 \\ & \therefore x=\frac{1}{2} \text { or } x=\frac{-2}{3}\end{aligned}$

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Question 71 Mark

Find $x$, if $g(x)=0$ where : $g(x)=\frac{18-2 x^2}{7}$

Answer

$\begin{aligned} & \mathrm{g}(\mathrm{x})=\frac{18-2 x^2}{7} \\ & \mathrm{~g}(\mathrm{x})=0 \\ & \therefore \frac{18-2 x^2}{7}=0 \\ & \therefore 18-2 \mathrm{x}^2=0 \\ & \therefore \mathrm{x}^2=9 \\ & \therefore \mathrm{x}= \pm 3\end{aligned}$

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Question 81 Mark

Find $x$, if $g(x)=0$ where : $g(x)=\frac{5 x-6}{7}$

Answer

$\begin{aligned} & \mathrm{g}(\mathrm{x})=\frac{5 x-6}{7} \\ & \mathrm{~g}(\mathrm{x})=0 \\ & \therefore \frac{5 x-6}{7}=0 \\ & \therefore 5 \mathrm{x}-6=0 \\ & \therefore \mathrm{x}=\frac{6}{5}\end{aligned}$

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Question 91 Mark

If $f(m)=m^2-3 m+1$, find : $f(-x)$

Answer

$f(-x)=(-x)^2-3(-x)+1=x^2+3 x+1$

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Question 101 Mark

If $f(m)=m^2-3 m+1$, find : $f(x+1)$

Answer

$
\begin{aligned}
& f(x+1)=(x+1)^2-3(x+1)+1 \\
& =x^2+2 x+1-3 x-3+1 \\
& =x^2-x-1
\end{aligned}
$

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Question 111 Mark

If $f(m)=m^2-3 m+1$, find : $f\left(\frac{1}{2}\right)$

Answer

$
\begin{aligned}
& f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2-3\left(\frac{1}{2}\right)+1 \\
& =\frac{1}{4}-\frac{3}{2}+1 \\
& =\frac{1-6+4}{4} \\
& =-\frac{1}{4} \\
&
\end{aligned}
$

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Question 121 Mark

If $f(m)=m^2-3 m+1$, find : $f(-3)$

Answer

$
\begin{aligned}
& f(-3)=(-3)^2-3(-3)+1 \\
& =9+9+1 \\
& =19
\end{aligned}
$

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Question 131 Mark

If $f(m)=m^2-3 m+1$, find : $f(0)$

Answer

$
\begin{aligned}
& f(m)=m^2-3 m+1 \\
& f(0)=0^2-3(0)+1=1
\end{aligned}
$

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Maths (commerce) Solve the following Question.(1 Marks)

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